Solving Quadratics by Factoring: Additional Practice (Set 2)

Its zeros are $\pm\sqrt{3}$, which are real but irrational, so a rational factorization is impossible while a real one is not.

$x^2 - 3 = \big(x - \sqrt{3}\big)\big(x + \sqrt{3}\big)$

The factorization uses irrational coefficients, so it factors over the reals but not over the rationals. "Does not factor" is meaningless until the number system is named.

Find two numbers with product $40$ and sum $-13$: those are $-5$ and $-8$.

$x^2 - 13x + 40 = (x - 5)(x - 8) = 0$

Then $x = 5$ or $x = 8$, giving the solution set $\{5, 8\}$.

Substitute the known point and solve for $a$.

$-8 = a(3 - 1)(3 - 5) = a(2)(-2) = -4a \ \Rightarrow \ a = 2$

The zeros $1$ and $5$ fix the factors but not the leading coefficient, so one extra point is needed to determine $a = 2$.

The right side is not zero, so expand and move all terms to one side first.

$x^2 - 4x - 5 = -5 \ \Rightarrow \ x^2 - 4x = x(x - 4) = 0$

Then $x = 0$ or $x = 4$. Setting factors equal to $-5$ directly would be invalid, since only a product of zero forces a factor. The solution set is $\{0, 4\}$.

A single solution means a perfect-square trinomial. Check $x^2 - 14x + 49$: the constant $49 = 7^2$ and $14x = 2 \cdot 7 \cdot x$.

$x^2 - 14x + 49 = (x - 7)^2 = 0 \ \Rightarrow \ x = 7$

The others each have two distinct solutions ($\pm 2$, then $2$ and $3$, then $\pm 3$).

Tangent to the axis means a repeated root at the point of tangency, $x = -2$.

$y = (x - (-2))^2 = (x + 2)^2$

Expanding gives $x^2 + 4x + 4$, which has the single zero $x = -2$ and vertex $(-2, 0)$ on the axis.

The Zero Product Property joins the factor conditions with "or," and "or" corresponds to the union of solution sets.

$x = 1 \ \text{ or } \ x = 2 \ \text{ or } \ x = 3$

So the solution set is $\{1\} \cup \{2\} \cup \{3\} = \{1, 2, 3\}$. An intersection would be empty, since no single number is all three at once.

Bring every term to one side so a product equals zero, then factor.

$x^2 - 7x + 12 - x + 3 = x^2 - 8x + 15 = (x - 3)(x - 5) = 0$

Then $x = 3$ or $x = 5$, giving the solution set $\{3, 5\}$.

Factor to find the roots, then multiply.

$x^2 - 7x + 10 = (x - 2)(x - 5) = 0 \ \Rightarrow \ x = 2, \ x = 5$

Their product is $2 \cdot 5 = 10$, which matches the constant term $c = 10$ for a monic quadratic.

Factor out the common $3x$; do not divide by $x$.

$3x^2 - 27x = 3x(x - 9) = 0$

The factor $3$ is never zero, so $x = 0$ or $x = 9$. The solution set is $\{0, 9\}$.

Let $u = x^2$, so the equation becomes $u^2 - 5u + 4 = 0$, which factors.

$u^2 - 5u + 4 = (u - 1)(u - 4) = 0 \ \Rightarrow \ x^2 = 1 \ \text{ or } \ x^2 = 4$

Then $x = \pm 1$ or $x = \pm 2$, four real solutions in all: $-2, -1, 1, 2$.

Build the factored form from the zeros and expand.

$(x - 2)(x + 9) = x^2 + 9x - 2x - 18 = x^2 + 7x - 18$

Matching $x^2 + bx + c$ gives $b = 7$ and $c = -18$. (The sum of the zeros is $-b = -7$ and their product is $c = -18$.)